 # A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls.

Question:

A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is

(i) black,
(ii) not green,
(iii) red or white,
(iv) neither red nor green.

Solution:

Total number of balls = 15

(i) Number of black balls = 2

$\therefore P($ getting a black ball $)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$

$=\frac{2}{15}$

Thus, the probability of getting a black ball is $\frac{2}{15}$.

(ii) Number of balls which are not green = 4 + 5 + 2 = 11

$\therefore \mathrm{P}$ (getting a ball which is not green) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$

$=\frac{11}{15}$

Thus, the probability of getting a ball which is not green is $\frac{11}{15}$.

(iii) Number of balls which are either red or white = 4 + 5 = 9

$\therefore P$ (getting a ball which is red or white) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$

$=\frac{9}{15}=\frac{3}{5}$

Thus, the probability of getting a ball which is red or white is $\frac{3}{5}$.

(iv) Number of balls which are neither red nor green = 4 + 2 = 6

$\therefore \mathrm{P}$ (getting a ball which is neither red nor green) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$

$=\frac{6}{15}=\frac{2}{5}$

Thus, the probability of getting a ball which is neither red nor green is $\frac{2}{5}$.